A 120 kg motorbike plus 80 kg rider is to accelerate from zero to 5 m

1.	A 120 kg motorbike plus 80 kg rider is to accelerate from zero to 5 m S^-2 in 0.8 S. Estimate the average force required.
Answer» Explanation: m = 120kg + 80kg = 200kg A1 = 0m/s² A2 = 5m/s² t = 0.8sec F = ma ( Newton's second LAW of motion) F1 = (200)(0) = 0N ----(1) F2 = (200)(5) =. 1000N ----(2) now, (1)+(2)/2 -----> average = 0+1000/2 = 1000/2 = 500N ...:)!):) might help you

A 120 kg motorbike plus 80 kg rider is to accelerate from zero to 5 m S^-2 in 0.8 S. Estimate the average force required.

Answer»

Explanation:

m = 120kg + 80kg = 200kg

A1 = 0m/s²

A2 = 5m/s²

t = 0.8sec

F = ma ( Newton's second LAW of motion)

F1 = (200)(0)

= 0N ----(1)

F2 = (200)(5)

=. 1000N ----(2)

now, (1)+(2)/2 -----> average

= 0+1000/2

= 1000/2 = 500N ...:)!):) might help you

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A 120 kg motorbike plus 80 kg rider is to accelerate from zero to 5 m S^-2 in 0.8 S. Estimate the average force required.​

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A 120 kg motorbike plus 80 kg rider is to accelerate from zero to 5 m S^-2 in 0.8 S. Estimate the average force required.